Question: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.1$ years; the standard deviation is $1.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between $16.9$ and $18.8$ years.
$13.1$ $11.2$ $15$ $9.3$ $16.9$ $7.4$ $18.8$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $13.1$ years. We know the standard deviation is $1.9$ years, so one standard deviation below the mean is $11.2$ years and one standard deviation above the mean is $15$ years. Two standard deviations below the mean is $9.3$ years and two standard deviations above the mean is $16.9$ years. Three standard deviations below the mean is $7.4$ years and three standard deviations above the mean is $18.8$ years. We are interested in the probability of a lion living between $16.9$ and $18.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of lions between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular lion living between $16.9$ and $18.8$ years is $\color{orange}{2.35\%}$.